What is the answer to z^2=2+2isqrt(3) ?

The answer to z^2=2+2isqrt(3) is z=sqrt(3)+i,z=-sqrt(3)-i

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z^2=2+2isqrt(3)

z^{2} = 2 + 2 i 3

z = 3 + i, z = - 3 - i

Solution steps

z^{2} = 2 + 2 i 3

Substitute

z = x + y i

(x + y i)^{2} = 2 + 2 i 3

Expand

(x + y i)^{2} : (x^{2} - y^{2}) + 2 i x y

(x^{2} - y^{2}) + 2 i x y = 2 + 2 i 3

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[x^{2} - y^{2} = 2 2 x y = 23]

[x^{2} - y^{2} = 2 2 x y = 23] : (x = 3, x = - 3, y = 1 y = - 1)

Substitute back

z = x + y i

z = 3 + i, z = - 3 - i

x^{2} - 3 i x - 2 = 0

6 + (x + 3 y) i = x + 3 i

x (1 + 3 i) + y (2 - i) - 5 + i^{3} = 0

z_{1} + i z_{2} = 1

x^{2} = - 2 - 23 i

What is the answer to z^2=2+2isqrt(3) ?
The answer to z^2=2+2isqrt(3) is z=sqrt(3)+i,z=-sqrt(3)-i